At a point on a horizontal line through the base of a monument, the angle of elevation of the top of the monument is found to be such that its tangent is $\dfrac {1}{5}$ . On walking 138 metres towards the monument the secant of the angle of elevation is found to be $\dfrac {\sqrt{193}}{12}$. The height of the monument (in metre) is
Solution
$\tan \theta = \dfrac{1}{5} = \dfrac{AC}{BC}$
$\dfrac{AC}{BC} = \dfrac{1}{5}$
BD = 138
$\sec \alpha = \dfrac{AD}{AC}$
$\dfrac{AD}{DC} = \dfrac{193}{12} \Rightarrow AC = \sqrt{193 - 144} = 7$
If AC = 7 then BC = 35
BD = BC - DC $\Rightarrow 35 - 12 = 138 \Rightarrow 1$ unit = 6
AC $= 7 \times 6 = 42$m